如何找到共同的元素在陣列重複紅寶石

Question

比方說陣列看起來像下面

city = ['london', 'new york', 'london', 'london', 'washington'] 

desired_location = ['london'] 

city & desired_location給['london']

，但我想['london', 'london', 'london']

Answer 1

您可以使用Enumerable#select

city.select {|c| desired_location.include?(c)} 
# => ["london", "london", "london"] 

Answer 2

另一種方法：

cities = ['london', 'new york', 'london', 'london', 'washington'] 
puts cities.select{|city| cities.count(city) > 1} 

Answer 3

cities = ['london', 'new york', 'london', 'london', 'washington'] 

如果desired_location包含一個元素：

desired_location = ['london'] 

我建議@桑托斯的解決方案，但是這也適用：

desired_location.flat_map { |c| [c]*cities.count(c) } 
    #=> ["london", "london", "london"] 

假設desired_location包含多個元素（這是我假設是一種可能性，否則將不需要它是一個數組）：

desired_location = ['london', 'new york'] 

@Santosh」方法的返回值：

["london", "new York", "london", "london"] 

這很可能是你想要的。如果你喜歡他們進行分組：

desired_location.flat_map { |c| [c]*cities.count(c) } 
    #=> ["london", "london", "london", "new york"] 

或：

desired_location.map { |c| [c]*cities.count(c) } 
    #=> [["london", "london", "london"], ["new york"]] 

根據您的要求，您可能會發現它更適用於生產哈希：

Hash[desired_location.map { |c| [c, cities.count(c)] }] 
    #=> {"london"=>3, "new york"=>1}